Let F be the force applied to separate the plates of a parallel plate capacitor by a distance of △x. Hence, work done by the force to do so =F△x
As a result, the potential energy of the capacitor increases by an amount given as uA△x
Where,
u= Energy density
A= Area of each plate
d= Distance between the plates
V= Potential difference across the plates
The work done will be equal to the increase in the potential energy i.e.,
F△x=uA△x
F=uA=(12∈0E2)A
Electric intensity is given by,
E=Vd
∴F=12∈0(Vd)EA=12(∈0AVd)E
However, capacitance, C=∈0Ad
∴F=12(CV)E
Charge on the capacitor is given by,
Q=CV
∴F=12QE
The physical origin of the factor, 12, in the force formula lies in the fact that just outside the conductor, field is E and inside it is zero. Hence, it is the average value, E2, of the field that contributes to the force.